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**Hint:**In the given problem, we are required to differentiate the given composite function layer by layer using the chain rule of differentiation. The basic power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{\left( {n - 1} \right)}}$ and the chain rule of differentiation are to be used in the given question so as to differentiate the given composite function. We need to keep this in mind the derivatives of basic functions such as logarithmic function in order to solve the problem.

**Complete step by step solution:**

Here we have a nested composite function, so we apply chain rule of differentiation. Here, we take $f\left( {g\left( {p\left( x \right)} \right)} \right)$ as \[{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)^2}\]. So, we first differentiate \[{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)^2}\] with respect to \[\left( {\ln \left( {\tan \left( x \right)} \right)} \right)\], then differentiate \[\ln \left( {\tan \left( x \right)} \right)\] with respect to \[\tan (x)\]and at last, differentiate \[\tan (x)\]with respect to x. When we differentiate \[{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)^2}\] with respect to \[\ln \left( {\tan \left( x \right)} \right)\], we get \[2\ln \left( {\tan \left( x \right)} \right)\], when we differentiate \[\ln \left( {\tan \left( x \right)} \right)\] with respect to \[\tan \left( x \right)\], we get \[\dfrac{1}{{\tan \left( x \right)}}\] and when we differentiate \[\tan \left( x \right)\] with respect to x, we get \[{\sec ^2}\left( x \right)\].

The formula for chain rule is given by,

$\dfrac{d}{{dx}}\left[ {f\left( {g\left( {p\left( x \right)} \right)} \right)} \right] = \dfrac{d}{{d\left( {g\left( {p\left( x \right)} \right)} \right)}}\left[ {f\left( {g\left( {p\left( x \right)} \right)} \right)} \right] \times \dfrac{d}{{d\left( {p\left( x \right)} \right)}}\left[ {g\left( {p\left( x \right)} \right)} \right] \times \dfrac{{d\left( {p\left( x \right)} \right)}}{{dx}}$

Putting in the function given to us and differentiating it layer by layer, we get,

So, $\dfrac{d}{{dx}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] = \dfrac{d}{{d\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] \times \dfrac{d}{{d\left( {\tan x} \right)}}\left[ {\ln \left( {\tan \left( x \right)} \right)} \right] \times \dfrac{{d\left( {\tan x} \right)}}{{dx}}$

On substituting all the derivatives, we get,

\[\dfrac{d}{{dx}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] = 2\ln \left( {\tan \left( x \right)} \right) \times \dfrac{1}{{\tan x}} \times {\sec ^2}x\]

Rewriting $\tan x$ as $\left( {\dfrac{{\sin x}}{{\cos x}}} \right)$, we get,

\[\dfrac{d}{{dx}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] = 2\ln \left( {\tan \left( x \right)} \right) \times \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}}} \times \dfrac{1}{{{{\cos }^2}x}}\]

\[\dfrac{d}{{dx}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] = 2\ln \left( {\tan \left( x \right)} \right) \times \dfrac{1}{{\sin x}} \times \dfrac{1}{{\cos x}}\]

Simplifying further,

\[\dfrac{d}{{dx}}\left[ {{{\left( {\ln \left( {\tan \left( x \right)} \right)} \right)}^2}} \right] = 2\dfrac{{\ln \left( {\tan \left( x \right)} \right)}}{{\sin x\cos x}}\]

**This is our required solution.**

**Note:**This chain rule is applied, when there is function of function in the given equation. It should be kept in mind forever, because it is important to know this formula to solve many complex equations. \[f(x) = {\left( {\ln \left( {\tan \left( x \right)} \right)} \right)^2}\] is mentioned as function of function or composite function. Here in this problem, it has a logarithmic function to base e which is represented by \[f(x) = {\left( {\ln \left( {\tan \left( x \right)} \right)} \right)^2}\] . The difference between $\ln $ and $\log $ is, in natural logarithm, $\ln $ ,it has the base e in it, while in the $\log $, it has the base $10$. $\log $ tells you that what power does $10$ has to be raised to get a number x and $\ln $ tells us that what power does we have to be raised to get a number x.

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